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\centerline{\Kopfont Commutative Algebra}

\smallskip
\centerline{Excercises 4}


\bigskip\noindent
Let $\phi: A \to B$ be a homomoprhism of rings. We say
that the {\it going-up theorem} holds for $\phi$ if the
following condition is satisfied:
\itemitem{(GU)} for any $\gp, \gp' \in \Spec(A)$ such that
$\gp \subset \gp'$, and for any $P \in \Spec(B)$ lying
over $\gp$, there exists $P'\in \Spec(B)$ lying
over $\gp'$ such that $P \subset P'$.

\noindent
Similarly, we say that the {\it going-down theorem} holds for $\phi$
if the following condition is satisfied:
\itemitem{(GD)} for any $\gp, \gp' \in \Spec(A)$ such that
$\gp \subset \gp'$, and for any $P' \in \Spec(B)$ lying
over $\gp'$, there exists $P\in \Spec(B)$ lying
over $\gp$ such that $P \subset P'$.

\medskip\noindent
{\bf 1.} In each of the following cases determine whether
(GU), (GD) holds, and explain why. (Use any Prop/Thm/Lemma you can find, 
but check the hypotheses in each case.)
\item{(a)} $k$ is a field, $A=k$, $B=k[x]$.
\item{(b)} $k$ is a field, $A=k[x]$, $B=k[x,y]$.
\item{(c)} $A={\bbb Z}$, $B={\bbb Z}[1/11]$.
\item{(d)} $k$ is an algebraically closed field, $A=k[x,y]$, $B=k[x,y,z]/(x^2-y,z^2-x)$.
\item{(e)} $A={\bbb Z}$, $B={\bbb Z}[i,1/(2+i)]$.
\item{(f)} $A={\bbb Z}$, $B={\bbb Z}[i,1/(14+7i)]$.
\item{(g)} $k$ is an algebraically closed field, $A=k[x]$, $B=k[x,y,1/(xy-1)]/(y^2-y)$.

\medskip\noindent
{\bf 2.} Let $k$ be an algebraically closed field. Compute the image in $\Spec(k[x,y])$
of the following maps:
\item{(a)} $\Spec(k[x,yx^{-1}]) \to \Spec(k[x,y])$, where
$k[x,y] \subset k[x,yx^{-1}] \subset k[x,y,x^{-1}]$.
(Hint: To avoid confusion, give the element $yx^{-1}$ another name.)
\item{(b)} $\Spec(k[x,y,a,b]/(ax-by-1))\to \Spec(k[x,y])$.
\item{(c)} $\Spec(k[t,1/(t-1)]) \to \Spec(k[x,y])$, induced by $x \mapsto t^2$,
and $y \mapsto t^3$.
\item{(d)} $k={\bbb C}$ (complex numbers),
$\Spec(k[s,t]/(s^3+t^3-1)) \to \Spec(k[x,y])$, where
$x\mapsto s^2$, $y \mapsto t^2$.

\medskip\noindent
{\bf Remark.} Finding the image as above usually is done by using
elimination theory.

\medskip\noindent
{\bf 3.} Let $k$ be a field. Show that the following pairs of 
$k$-algebras are not isomorphic:
\item{(a)} $k[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_{n+1}]$ for any
$n\geq 1$.
\item{(b)} $k[a,b,c,d,e,f]/(ab+cd+ef)$ and $k[x_1,\ldots,x_n]$ 
for $n=5$.
\item{(c)} $k[a,b,c,d,e,f]/(ab+cd+ef)$ and $k[x_1,\ldots,x_n]$ 
for $n=6$.

\medskip\noindent
{\bf Remark.} Of course the idea of this exercise is to find 
a simple argument in each case rather than applying a ``big'' theorem.
Nonetheless it is good to be guided by general principles.

\end
